// Numbas version: finer_feedback_settings {"name": "Gareth's copy of Differentiation using the chain rule", "feedback": {"showtotalmark": true, "advicethreshold": 0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true, "enterreviewmodeimmediately": true, "showexpectedanswerswhen": "inreview", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "never"}, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "allQuestions": true, "shuffleQuestions": false, "percentPass": 0, "duration": 0, "pickQuestions": 0, "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "showresultspage": "oncompletion", "preventleave": true, "browse": true, "showfrontpage": true}, "metadata": {"description": "

Use the chain rule to differentiate various functions.

", "licence": "Creative Commons Attribution 4.0 International"}, "type": "exam", "questions": [], "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "Chain rule - binomial", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "tags": ["Calculus", "calculus", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "Differentiation", "function of a function", "Steps", "steps"], "metadata": {"description": "

Differentiate $\\displaystyle (ax^m+b)^{n}$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Differentiate the following function $f(x)$ using the chain rule.

", "advice": "

$\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}$

The chain rule says that if $f(x)=g(h(x))$ then

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

Then we use the chain rule in the form:

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u}\\]

Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

For this example, we let $u=\\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\\simplify[std]{u^{n}}$.

This gives

\\begin{align}
\\frac{\\mathrm{d}u}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m -1}} \\\\[1em]
\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u} &= \\simplify[std]{{n}u^{n-1}}
\\end{align}

Hence on substituting into the chain rule above we get:

\\begin{align}
\\frac{\\mathrm{d}f}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m-1} * ({n}*u^{n-1})} \\\\
&= \\simplify[std]{{m*a*n}x^{m-1}u^{n-1}} \\\\
&= \\simplify[std]{{m*a*n}x^{m-1}({a}*x^{m}+{b})^{n-1}}
\\end{align}

on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.

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\\[\\simplify[std]{f(x) = ({a} * x^{m}+{b})^{n}}\\]

$\\displaystyle \\frac{\\mathrm{d}f}{\\mathrm{d}x}=$ [[0]]

Click on Show steps for more information. You will not lose any marks by doing so.

", "stepsPenalty": 0, "steps": [{"type": "information", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

The chain rule says that if $f(x)=g(h(x))$ then

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

Then we use the chain rule in the form:

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f}{\\mathrm{d}u}\\]

Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

"}], "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 3, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "answer": "{a*m*n}x ^ {m-1} * ({a} * x^{m}+{b})^{n-1}", "answerSimplification": "std", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": true, "singleLetterVariables": false, "allowUnknownFunctions": false, "implicitFunctionComposition": false, "valuegenerators": [{"name": "x", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Chain rule - exponential of polynomial, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"s1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s1"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "b"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s2*random(1..9)", "description": "", "name": "c"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "a"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s2"}, "m": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(3..4)", "description": "", "name": "m"}}, "ungrouped_variables": ["a", "c", "b", "s2", "s1", "m"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 0, "scripts": {}, "gaps": [{"answer": "({m*a}x^{m-1}+{2*b}x)*e^({a}x^{m} +{b}x^2+{c})", "showCorrectAnswer": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t

\\[\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}\\]

\n\t\t\t

$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

\n\t\t\t

Click on Show steps for more information. You will not lose any marks by doing so.

\n\t\t\t", "steps": [{"type": "information", "prompt": "\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t

The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "

Differentiate the following function $f(x)$ using the chain rule.

", "tags": ["Calculus", "MAS1601", "SFY0004", "Steps", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t

1/08/2012:

\n\t\t

Added tags.

\n\t\t

Added description.

\n\t\t

Checked calculation. OK.

\n\t\t

Added information about Show steps. Altered to 0 marks lost rather than 1.

\n\t\t

Got rid of a redundant ruleset.

\n\t\t

Improved display in prompt.

\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "

Differentiate $\\displaystyle e^{ax^{m} +bx^2+c}$

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t \n\t \n\t

$\\simplify[std]{f(x) = e^({a}x^{m} +{b}x^2+{c})}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

\n\t \n\t \n\t \n\t

For this example, we let $u=\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{e^u}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a*m}x^{m-1} +{2*b}x}\\\\\n\t \n\t \\frac{df(u)}{du} &=& \\simplify[std]{e^u} \\end{eqnarray*}\\]

\n\t \n\t \n\t \n\t

Hence on substituting into the chain rule above we get:

\n\t \n\t \n\t \n\t

\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x) * (e^u)}\\\\\n\t \n\t &=& \\simplify[std]{({a*m}x^{m-1} +{2*b}x)*e^({a}x^{m} +{b}x^2+{c})}\n\t \n\t \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m} +{b}x^2+{c}}$.

\n\t \n\t \n\t"}, {"name": "Chain rule - log of binomial", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"s1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s1"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "b"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s2*random(1..9)", "description": "", "name": "c"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "a"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s2"}, "m": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(3..9)", "description": "", "name": "m"}}, "ungrouped_variables": ["a", "c", "b", "s2", "s1", "m"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 0, "scripts": {}, "gaps": [{"answer": "({m*a})/({a}x+{b})", "showCorrectAnswer": true, "vsetrange": [5, 6], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "\n\t\t\t

\\[\\simplify[std]{f(x) = ln(({a}x+{b})^{m})}\\]

\n\t\t\t

$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

\n\t\t\t

Click on Show steps for more information. You will not lose any marks by doing so.

\n\t\t\t", "steps": [{"type": "information", "prompt": "\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t

\n\t\t\t\t\t \n\t\t\t\t\t \n\t\t\t\t\t", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "

Differentiate the following function $f(x)$ using the chain rule.

", "tags": ["Calculus", "MAS1601", "Steps", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function", "logarithm laws", "logarithms"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "\n\t\t

\n\t\t

1/08/2012:

\n\t\t

Added tags.

\n\t\t

Added description.

\n\t\t

Checked calculation. OK.

\n\t\t

Added information about Show steps. Altered to 0 marks lost rather than 1.

\n\t\t

Got rid of a redundant ruleset.

\n\t\t

Improved display in prompt.

\n\t\t

Checking range chosen so that the denominator of the result is never 0.

\n\t\t

\n\t\t", "licence": "Creative Commons Attribution 4.0 International", "description": "

Differentiate $\\displaystyle \\ln((ax+b)^{m})$

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n\t \n\t \n\t

$\\simplify[std]{f(x) = ln(({a}x+{b})^{m})}$
First note that we can simplify this by using the rule that $\\simplify[std]{ln(a^r)=r*ln(a)}$.
Hence $\\simplify[std]{f(x) = ln(({a}x+{b})^{m})={m}ln({a}x+{b})}$
So we need to differentiate $\\simplify[std]{ln({a}x+{b})}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

\n\t \n\t \n\t \n\t

For this example, we let $u=\\simplify[std]{{a}x +{b}}$ and we have $f(u)=\\simplify[std]{{m}*ln(u)}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{a}}\\\\\n\t \n\t \\frac{df(u)}{du} &=& \\simplify[std]{{m}/u} \\end{eqnarray*}\\]

\n\t \n\t \n\t \n\t

Hence on substituting into the chain rule above we get:

\n\t \n\t \n\t \n\t

\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({a}) * ({m}/u)}\\\\\n\t \n\t &=& \\simplify[std]{{a*m}/({a}x+{b})}\n\t \n\t \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x+{b}}$.

\n\t \n\t \n\t"}, {"name": "Chain rule - power of a quadratic, ", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "variable_groups": [], "variables": {"s1": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s1"}, "n": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(3..5)", "description": "", "name": "n"}, "b": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s1*random(1..9)", "description": "", "name": "b"}, "c": {"templateType": "anything", "group": "Ungrouped variables", "definition": "s2*random(1..9)", "description": "", "name": "c"}, "a": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "name": "a"}, "s2": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "name": "s2"}, "m": {"templateType": "anything", "group": "Ungrouped variables", "definition": "random(3..7)", "description": "", "name": "m"}}, "ungrouped_variables": ["a", "c", "b", "s2", "s1", "m", "n"], "question_groups": [{"pickingStrategy": "all-ordered", "questions": [], "name": "", "pickQuestions": 0}], "functions": {}, "showQuestionGroupNames": false, "parts": [{"stepsPenalty": 0, "scripts": {}, "gaps": [{"answer": "{n}({a*m}x ^ {m-1}+{2*c}x) * ({a} * x^{m}+{c}x^2+{b})^{n-1}", "showCorrectAnswer": true, "vsetrange": [0, 1], "checkingaccuracy": 0.001, "checkvariablenames": false, "expectedvariablenames": [], "showpreview": true, "checkingtype": "absdiff", "scripts": {}, "type": "jme", "answersimplification": "std", "marks": 3, "vsetrangepoints": 5}], "type": "gapfill", "prompt": "

\\[\\simplify[std]{f(x) = ({a} * x^{m}+{c}x^2+{b})^{n}}\\]

$\\displaystyle \\frac{df}{dx}=\\;$[[0]]

Click on Show steps for more information. You will not lose any marks by doing so.

", "steps": [{"type": "information", "prompt": "\n \n \n

\n \n ", "showCorrectAnswer": true, "scripts": {}, "marks": 0}], "showCorrectAnswer": true, "marks": 0}], "statement": "

Differentiate the following function $f(x)$ using the chain rule.

", "tags": ["Calculus", "MAS1601", "SFY0004", "chain rule", "checked2015", "derivative of a function of a function", "differentiation", "function of a function"], "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "preamble": {"css": "", "js": ""}, "type": "question", "metadata": {"notes": "

1/08/2012:

Added tags.

Added description.

Checked calculation. OK.

Added information about Show steps. Altered to 0 marks lost rather than 1.

Got rid of a redundant ruleset.

Improved display in prompt.

", "licence": "Creative Commons Attribution 4.0 International", "description": "

Differentiate $\\displaystyle (ax^m+bx^2+c)^{n}$.

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "advice": "\n \n \n

$\\simplify[std]{f(x) = f(x) = ({a} * x^{m}+{c}x^2+{b})^{n}}$
The chain rule says that if $f(x)=g(h(x))$ then
\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]
One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.
Then we use the chain rule in the form:
\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]
Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

\n \n \n \n

For this example, we let $u=\\simplify[std]{{a} * x^{m}+{c}*x^2+{b}}$ and we have $f(u)=\\simplify[std]{u^{n}}$.
This gives
\\[\\begin{eqnarray*}\\frac{du}{dx} &=& \\simplify[std]{{m*a}x ^ {m -1}+{2*c}x}\\\\\n \n \\frac{df(u)}{du} &=& \\simplify[std]{{n}u^{n-1}} \\end{eqnarray*}\\]

\n \n \n \n

Hence on substituting into the chain rule above we get:

\n \n \n \n

\\[\\begin{eqnarray*}\\frac{df}{dx} &=& \\simplify[std]{({m*a}x ^ {m-1}+{2*c}x) * ({n}*u^{n-1})}\\\\\n \n &=&\\simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)u^{n-1}}\\\\\n \n &=& \\simplify[std]{{n}*({m*a}x^{m-1}+{2*c}*x)({a}*x^{m}+{c}*x^2+{b})^{n-1}}\n \n \\end{eqnarray*}\\]
on replacing $u$ by $\\simplify[std]{{a}x^{m}+{c}x^2+{b}}$.

\n \n "}, {"name": "Chain rule - product of two functions", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}], "tags": ["algebraic manipulation", "Calculus", "calculus", "chain rule", "Chain rule", "checked2015", "derivative of the product of two functions", "differentiation", "Differentiation", "product rule"], "metadata": {"description": "

The derivative of $\\displaystyle x ^ {m}(ax^2+b)^{n}$ is of the form $\\displaystyle x^{m-1}(ax^2+b)^{n-1}g(x)$. Find $g(x)$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

$\\simplify[std]{f(x) = x ^ {m} * ({a} * x^2+{b})^{n}}$

", "advice": "

The product rule says that if $u$ and $v$ are functions of $x$ then

\\[\\simplify[std]{Diff(u * v,x,1) = u * Diff(v,x,1) + v * Diff(u,x,1)}\\]

For this example:

\\[\\simplify[std]{u = x ^ {m}}\\Rightarrow \\simplify[std]{Diff(u,x,1) = {m}x ^ {m -1}}\\]

\\[\\simplify[std]{v = ({a} * x^2+{b})^{n}} \\Rightarrow \\simplify[std]{Diff(v,x,1) = {2*n*a}*x * ({a} * x^2+{b})^{n-1}}\\]

For this last differentiation we used the chain rule.

Hence on substituting into the product rule above we get:

\\begin{align}
\\frac{\\mathrm{d}f}{\\mathrm{d}x} &= \\simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+x^{m} *{2*n*a}*x* ({a} * x^2+{b})^{n-1}}\\\\
&= \\simplify[std]{{m}x ^ {m-1} * ({a} * x^2+{b})^{n}+{2*n*a}*x^{m+1}* ({a} * x^2+{b})^{n-1}}\\\\
&= \\simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m}*({a}*x^2+{b})+{2*n*a}x^{2})} \\\\
&= \\simplify[std]{x ^ {m-1} * ({a} * x^2+{b})^{n-1}*({m*a+2*a*n}*x^2+{m*b})}
\\end{align}

Hence $\\simplify[std]{g(x)={m*a+2*a*n}*x^2+{m*b}}$

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Differentiate $f(x)$.

The answer can be written in the form

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x}=\\simplify[std]{x^{m-1}({a}x^2+{b})^{n-1}*g(x)}\\] for a polynomial $g(x)$.

You have to find $g(x)$.

$g(x)=$ [[0]]

Click on Show steps for more information.

You should use the the product rule and the chain rule for this example.

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Differentiate

\n\t\t

\\[ \\sqrt{a x^m+b})\\]

\n\t\t", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

Differentiate the following function $f(x)$ using the chain rule.

", "advice": "

$\\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}$

The chain rule says that if $f(x)=g(h(x))$ then

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

Then we use the chain rule in the form:

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u}\\]

Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

For this example, we let $u=\\simplify[std]{{a} * x^{m}+{b}}$ and we have $f(u)=\\simplify[std]{sqrt(u)=u^{1/2}}$.

This gives

\\begin{align}
\\frac{\\mathrm{d}u}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m -1}} \\\\[1em]
\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u} &= \\simplify[std]{{1/2}*u^{-1/2}=1/(2*sqrt(u))}
\\end{align}

Hence on substituting into the chain rule above we get:

\\begin{align}
\\frac{\\mathrm{d}f}{\\mathrm{d}x} &= \\simplify[std]{{m*a}x ^ {m-1} * (1/(2*sqrt(u)))} \\\\
&= \\simplify[std]{{m*a}x^{m-1}/(2*sqrt(u))} \\\\
&= \\simplify[std]{({a*m}x ^ {m-1})/(2*sqrt({a} * x^{m}+{b}))}
\\end{align}

on replacing $u$ by $\\simplify[std]{{a}x^{m}+{b}}$.

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\\[\\simplify[std]{f(x) = sqrt({a} * x^{m}+{b})}\\]

$\\displaystyle \\frac{\\mathrm{d}f}{\\mathrm{d}x}=$ [[0]]

Click on Show steps for more information.

The chain rule says that if $f(x)=g(h(x))$ then

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

Then we use the chain rule in the form:

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f}{\\mathrm{d}u}\\]

Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

Differentiate $\\displaystyle \\cos(e^{ax}+bx^2+c)$

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

$\\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}$

The chain rule says that if $f(x)=g(h(x))$ then

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

Then we use the chain rule in the form:

\\[\\frac{df}{dx} = \\frac{du}{dx}\\frac{df(u)}{du}\\]

Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.

For this example, we let $u=\\simplify[std]{e^({a}x) +{b}x^2+{c}}$ and we have $f(u)=\\simplify[std]{cos(u)}$.

This gives

\\begin{align}
\\frac{\\mathrm{d}u}{\\mathrm{d}x} &= \\simplify[std]{{a}e^({a}x) +{2*b}x} \\\\[1em]
\\frac{\\mathrm{d}f(u)}{\\mathrm{d}u} &= \\simplify[std]{-sin(u)}
\\end{align}

Hence on substituting into the chain rule above we get:

\\begin{align}
\\frac{\\mathrm{d}f}{\\mathrm{d}x} &= \\simplify[std]{({a}e^({a}x) +{2*b}x) * (-sin(u))} \\\\
&= \\simplify[std]{-({a}e^({a}x) +{2*b}x)*sin(e^({a}x) +{b}x^2+{c})}
\\end{align}

on replacing $u$ by $\\simplify[std]{e^({a}x) +{b}x^2+{c}}$.

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Differentiate the following function $f(x)$ using the chain rule.

\\[\\simplify[std]{f(x) = cos(e^({a}x) +{b}x^2+{c})}\\]

$\\displaystyle \\frac{\\mathrm{d}f}{\\mathrm{d}x}=$ [[0]]

The chain rule says that if $f(x)=g(h(x))$ then

\\[\\simplify[std]{f'(x) = h'(x)g'(h(x))}\\]

One way to find $f'(x)$ is to let $u=h(x)$ then we have $f(u)=g(u)$ as a function of $u$.

Then we use the chain rule in the form:

\\[\\frac{\\mathrm{d}f}{\\mathrm{d}x} = \\frac{\\mathrm{d}u}{\\mathrm{d}x}\\frac{\\mathrm{d}f}{\\mathrm{d}u}\\]

Once you have worked this out, you replace $u$ by $h(x)$ and your answer is now in terms of $x$.